

A233655


Sum of parts power divisors of canonical representation of n (A233569).


3



1, 2, 4, 4, 9, 9, 11, 8, 17, 12, 26, 17, 26, 26, 26, 16, 33, 26, 48, 26, 45, 45, 63, 33, 48, 45, 63, 48, 63, 63, 57, 32, 65, 50, 92, 40, 97, 97, 115, 50, 97, 54, 120, 97, 120, 120, 140, 65, 92, 97, 115, 97, 120, 120, 140, 92, 115, 120, 140, 115, 140, 140, 120, 64
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OFFSET

1,2


COMMENTS

If the canonical representation of n is A233569(n)=(1)^k_1[*](10)^k_2[*]...[*](10...0)^k_t, where [*] means concatenation, then we say that a number (1)^r_1[*](10)^r_2[*]...[*](10...0)^r_t is a parts power divisor of canonical representation of n, iff all r_i<=k_i.
Note that, by agreement, (10...0)^0 means the absence of the corresponding part.


LINKS

Table of n, a(n) for n=1..64.


FORMULA

a((10...0[m zeros])^k) = 2^m/(2^(m+1)1)^2 * (2^((m+1)*(k+1))  1)  (k+1)*2^m/(2^(m+1)1). For example, a(101010)[here m=1,k=3] = 2/9*(2^81)  4*2/3 = 54.
Thus a(42)=54. What is a general formula for a(n)?


EXAMPLE

Since A233569(5)=6, then the canonical representation of 5 is (1)^1[*](10)^1 which has parts power divisors 0, (1)^1, (10)^1, (1)^1[*](10)^1. Converting to decimal, they are 0,1,2,6 with sum 9. So a(5)=9. Note that 6 is a parts power divisor of 5, but not a cdivisors of 5 (see comment in A124771).
Analogously, 12 = (1)^1[*](10)^0[*](100)^1 is a parts power divisor of 52 = (1)^1[*](10)^1[*](100)^1, but not a cdivisor of 52.


CROSSREFS

Cf. A233394, A233569, A124771.
Sequence in context: A256701 A340714 A292381 * A307325 A272196 A335057
Adjacent sequences: A233652 A233653 A233654 * A233656 A233657 A233658


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, Dec 14 2013


STATUS

approved



